f(x)=(5x)/(1+2x^2).

f'(4)=-75/289

f'(x)=(5-5x^2)/1+2x^2+x^4)
Use this equation of the tangent line to the curve y=5x/(1+x^2) at the point (4, 1.17647). The equation of the tangent line can be written in the form y=mx+b. m=-75/289. What is b?

To find the value of b in the equation of the tangent line, we need to substitute the coordinates of the point (4, 1.17647) into the equation y = mx + b.

Given:
m = -75/289
Point: (4, 1.17647)

Substituting these values:
1.17647 = (-75/289)(4) + b

Now we can solve for b.

Multiply -75/289 by 4:
1.17647 = -300/289 + b

To simplify, we need to add -300/289 to both sides:
1.17647 + 300/289 = b

To add the fractions, we need a common denominator:
(289*1.17647 + 300)/(289) = b

Calculating the numerator:
335.8/289 = b

Simplifying the fraction:
b = 1.16

Therefore, the value of b in the equation of the tangent line y = mx + b is 1.16.